Post on 29-May-2020
Lecture 6Sobolev Theory
Dilution Factor I
sinθc = R/r
ω∗ ≡ solid angle subtended by stellar disk
W ≡ ω∗/(4π) ≡ dilution factor
Dilution Factor II
ω∗ =
∫dΩ = 2π
∫ θc
0sinθ dθ
= 2π(−cosθ)∣∣∣θc0
= 2π [1− cosθc]
= 2π[1−
√1− (R/r)2
]W = ω∗/(4π)
= 1/2[1−
√1− (R/r)2
]W (R) = 1/2
W (r >> R) = 1/4(R/r)2
Flux from a Star I
sin(90− θ) = z/r = cosθ; sinθ = p/r
µ = cosθ =√
r2 − p2/r
Flux from a Star II
dµ = −1/2(r2 − p2)−1/22p/rdp
dµ = −(r2 − p2)−1/2p/rdp
F = 2π∫ π
0I cosθd(cosθ)
= −2π∫ 0
rI
√r2 − p2
rp
r√
r2 − p2dp
=2πr2
∫ r
0I p dp
Constant Direction Velocity SurfacesSurfaces of Constant Line-of-sight Velocity
Resonant Scattering Qualitative
Calculation of FluxResonant Scattering
Sobolev TheoryCastor, 1970
Mathematics IRadiation travels from positive to negative z, that is from left to rightSurfaces of constant velocity: vz = µv
I(r , µ) = I(p, z)
p2 + z2 = r2; z = −µ r
The transfer equation is:
dIν(p, z)
dz= κν(p, z)(Iν(p, z)− S(r))
Assume thatIν(rc , µ) = Ic for all ν, µ
Iν(p, z) =
∫ ∞z
S((p2 + z ′2)1/2)eτν(p,z)−τν(p,z′) dτν(p, z ′)
for p > rc or z > 0
Mathematics II
Iν(p, z) =
∫ (r2c−p2)1/2
zS((p2 + z ′2)1/2)eτν(p,z)−τν(p,z′) dτν(p, z ′)
+Iceτν(p,z)−τν(p,−(r2c−p2)1/2)) for p < rc , z < 0
τν(p, z) =
∫ z
−∞κν(p, z ′)dz ′
κν(p, z) = κo(r)φ(ν +ν0
cv(r)
rz − ν0
φ(x) is strongly peaked at x = 0 so
ν0
cv(r)
r(z0)z0 = −(ν − ν0)
ν0
cv((p2 + z2
o )1/2)
(p2 + z2o )1/2
z0 = −(ν − ν0)
Mathematics IIIThen let x be the argument of φ
x = ν − ν0 + ν0v0(r ′)
r ′z ′
∫ z
−∞φ(ν − ν0 + ν0
v0(r ′)r ′
z ′) dz ′ =
∫ x
−∞φ(x ′)
dz ′
dxdx
≈[∫ x
−∞φ(x ′) dx ′
]1/
dxdz ′|z0
dxdz ′
=∂
∂z(ν0
cv(r)
rz)
=ν0
c
[v(r)
∂z ′/(z ′2 + p2)1/2
∂z ′+
z ′
rdvdr
drdz ′
]
drdz
=zr
More Math
dxdz ′
=ν0
c
[v(r)
r+ z
d(v/r)
dz ′
]=
ν0
cv(r)
r
[1 + z(
v(r)
r)−1 d(v/r)
drdrdz ′
]=
ν0
cv(r)
r
[1 +
z2
r(v(r)
r)−1 d(v/r)
dr
]=
ν0
cv(r)
r
[1 +
z2
r(v(r)
r)−1 d(v/r)
dr
]=
ν0
cv(r)
r
[1 +
z2
r(v(r)
r)−1[1/r
d(v)
dr− v
r2 ]
]=
ν0
cv(r)
r
[1 +
z2
rrv
[1/rd(v)
dr− v
r2 ]
]
More Math I
dxdz ′
=ν0
cv(r)
r
[1 +
z2
r2 [r/vd(v)
dr− 1]
]=
ν0
cv(r)
r
[1 +
z2
r2 [d ln vd ln r
− 1]
]So Finally
τν(p, z) = κ0((p2 + z2)1/2)y(ν + ν0
cvr − ν0)
ν0c
vr
[1 +
z2
r2 [d ln vd ln r
− 1]
]−1
y(x) =
∫ x
−∞φ(x ′) dx ′
y(+∞) = 1y(−∞) = 0
τν(p, z) = τν(p,∞)y
More Math II
In the limit:
φ→ δ
y → Θ(z − z0)
τν(p,∞) =
πe2
mc (gf )lu(Nlgl− Nu
gu)
νoc
vr
[1 +
z2
r2 [d ln vd ln r
− 1]
]−1
Iν(p, z) =
∫ 1
y(z)S((p2 + z2)1/2)eτν(p,∞)(y(z)−y(z′))
τν(p,∞)dy(z ′)
p > rcorz > 0
(1)
More Math III
Iν(p, z) =
∫ y(−(r2c−p2)1/2)
y(z)S((p2 + z2)1/2)eτν(p,∞)(y(z)−y(z′))
τν(p,∞)dy(z ′)
+Iceτν(p,∞)(y(z)−y(−(r2c−p2)1/2)) p < rc , z < 0
Recall,y(z) = y(ν +
ν0
cvr
z − ν0)
and that if z > z0 the integrals nearly vanish. If z < z0 then the lowerlimit→ 0. In this case most of the contribution to Iν comes at z = z0.Thus we can pull S out of the integral and evaluate it at z = z0
Iν(p, z) = S((p2 + z20 )1/2)
(1− eτν(p,∞)(y(z)−1))
)p > rcorz > 0
More Math IV
Iν(p, z) = S((p2 + z20 )1/2)
(1− eτν(p,∞)(y(z)−y(z′))
)+Iceτν(p,∞)(y(z)−y(−(r2
c−p2)1/2)) p < rc , z < 0
Flux I
F = 4π∫ rc
0
[S((p2 + z2
0 )1/2)(1− e−τ(p,∞)y(−(r2
c−p2)1/2))
+Ice−τ(p,∞)y(−(r2c−p2)1/2)
]2πp dp
+4π∫ ∞
rc
[S((p2 + z2
0 )1/2(1− e−τ(p,∞))]
2πp dp
Define
Fc = 4π∫ rc
02πpIc dp = 4π2r2
c Ic
Flux II
Fν − Fc
Fc=
1r2c
∫ ∞0
S((p2 + z20 )1/2)
Ic
(1− e−τ(p,∞)
)2p dp
− 1r2c
∫ rc
0
[1− e−τ(p,∞)y(−(r2
c−p2)1/2)]
2p dp
− 1r2c
∫ rc
0
S((p2 + z20 )1/2
Ic
[e−τ(p,∞)y(−(r2
c−p2)1/2)
−e−τ(p,∞)]
2p dp
I Term 1: Emission from entire core as if it were transparentI Term 2: Radiation removed from continuum by material in front of
the coreI Radiation that would have come from the part of the envelope that
is occulted by the core
Flux III
y(−((r2c − p2)1/2) =
0 if ∆ν < 01 if ∆ν > 0
So absorption and occultation terms are images of each other aroundthe line center modulo the factor S/Ic in the occultation term.
Shell
Homologous Lines
SYNOW Assumptions I
S = WIc
W =12
[1−
√1− rc
r
2]
v ∝ r
so photons come into resonance with lines i.e., y = 0,1 and weassume τ ≡ τ(r) which is specified to be for example a power-law orexponential in SYNOW. Then the flux for z ≥ 0 is
F (z) = 2π∫ rc
0Icp dp +
∫ ∞rc
S(r)[1− e−τ(r)
]p dp
= πr2c Ic + 2π
∫ ∞rc
S(r)[1− e−τ(r)
]p dp
SYNOW Assumptions II
Then the flux for z < 0 is
F (z) = 2π∫ p0
0Icp dp + 2π
∫ ∞p0
S(r)[1− e−τ(r)
]p dp
+2π∫ rc
p0
Ice−τ(r)p dp
= πp20Ic +
∫ ∞rc
S(r)[1− e−τ(r)
]p dp
p0 =
√r2c − z2 for −rc < z < 0
0 for z ≤ −rc